Answer:
2/3
Explanation:
The possible outcome from the three designations is express as 3!
and 3!=3x2x1=6
Hence the 6 possible outcome are listed below
(A,B,C),(A,C,B),(B,A,C),(B,C,A),(C,A,B) and (C,B,A)
let assume A should be less than B and B less than C, i.e A<B<C
The question required us to solve for
P(min(A,B)<C). And this can be expressed as
P(min(A,B)<C)=P(A<C)+P(B<C)-P(AB<C).
From the order given above, there are 3 possible outcome for A to come before C
Hence P(A<C)=3/6=1/2
also there 3 possible outcome for B to come first before C
P(B<C)=3/6=1/2
Also, there 2 events for A,B to come before C
P(AB<C)=2/6=1/3
Hence in final
P(min(A,B)<C)=P(A<C)+P(B<C)-P(AB<C)=1/2+1/2-1/3=2/3
the probability that it is also smaller than the value on card C is 2/3